Lipids — fatty acids, triglycerides, phospholipids; saponification: Worked examples

Every symbol used below is earned first. Quick anchor:

The scenario matrix

Every lipid problem is one of these case classes. The rest of the page fills each cell.

Cell	Case class	What varies / the "sign" here	Example
A	Counting bonds / waters	how many ester bonds break or form	Ex 1
B	Saturated vs unsaturated	$d = 0$ vs $d > 0$ → melting-point sign	Ex 2
C	Molar-mass / mass yield	how much soap from how much fat	Ex 3
D	Degenerate input: a monoglyceride	only 1 ester (not 3)	Ex 4
E	Base swap: NaOH vs KOH	Na vs K → hard vs soft soap	Ex 5
F	Acid vs base hydrolysis	which product: acid or salt	Ex 6
G	Amphipathic geometry (word problem)	which way heads/tails point	Ex 7 (figure)
H	Limiting case: fully unsaturated	$d$ large → liquid, oxidation	Ex 8
I	Exam twist: iodine number	measuring $d$ by chemistry	Ex 9

The key "signs" in this subject are not $\pm$ numbers — they are structural switches: $d = 0$ vs $d > 0$ (solid vs liquid), acid vs base medium (free acid vs salt), 1 vs 3 esters (mono vs tri). We hit each switch explicitly.

Cell A — Counting bonds and waters

Worked example Example 1 (Cell A)

A chemist joins 1 glycerol with 3 fatty acids to build a triglyceride, then later fully saponifies it. (a) How many water molecules were released during the build? (b) How many NaOH molecules are consumed in the saponification?

Forecast: guess both numbers before reading on.

Count the –OH groups on glycerol. Glycerol is propane-1,2,3-triol: $CH_{2} OH-CHOH-CH_{2} OH$ → 3 hydroxyl groups. Why this step? Each –OH can grab one fatty-acid –COOH to form one ester bond, and each ester formed releases one water.
Build: 3 esters formed → 3 waters released. (part a = 3) Why this step? Ester formation is a condensation: acid's –OH + alcohol's –H leave as $H_{2} O$ , once per bond.
Saponify: 3 esters must break → 3 NaOH. (part b = 3) Why this step? Saponification cleaves every ester; each cleavage uses one NaOH to make one soap salt.

Verify: 3 esters ⇔ 3 waters out (building) ⇔ 3 NaOH in (breaking). The count of ester bonds is the single invariant — it equals 3 for any triglyceride. ✓

Cell B — Saturated vs unsaturated (the melting switch)

Worked example Example 2 (Cell B)

Three fatty acids: stearic $18 : 0$ , oleic $18 : 1$ , linolenic $18 : 3$ . All have 18 carbons. Rank their melting points from highest to lowest and justify.

Forecast: which one is the waxy solid, which is the runniest oil?

Read $d$ (double bonds) from each shorthand. $18 : 0 \Rightarrow d = 0$ ; $18 : 1 \Rightarrow d = 1$ ; $18 : 3 \Rightarrow d = 3$ . Why this step? Melting point is governed almost entirely by how well chains stack, and each cis double bond (chain bends to the same side — see the definition above) adds a kink that ruins stacking.
Link kinks → packing → Van der Waals Forces. $d = 0$ is straight → chains lie flat against each other → maximum contact → strongest van der Waals attraction → hardest to melt. More kinks → less contact → weaker attraction. Why this step? Van der Waals force strength scales with contact area; a kink props chains apart.
Rank. Melting point: $18 : 0 > 18 : 1 > 18 : 3$ . So stearic (solid), oleic (liquid oil), linolenic (very fluid oil).

Verify: Real data — stearic ≈ 69 °C, oleic ≈ 13 °C, linolenic ≈ −11 °C. Order matches $d$ increasing → melting point decreasing. The sign of the switch ( $d = 0$ vs $d > 0$ ) correctly predicts solid vs liquid. ✓

Cell C — Molar mass and soap yield

Worked example Example 3 (Cell C)

How many grams of sodium stearate (soap) are produced by saponifying 445 g of pure tristearin with excess NaOH? Tristearin has molar mass $890 g/mol$ ; sodium stearate is $C_{17} H_{35} COONa$ .

Forecast: more or less than 445 g? (Trick: three sodiums are being added…)

Moles of tristearin. $\frac{445 g}{890 g/mol} = 0.5 mol$ . Why this step? Reactions count in moles, not grams; we must convert first.
Stoichiometry: 1 tristearin → 3 soap. So $0.5 \times 3 = 1.5 mol$ sodium stearate. Why this step? Each of the 3 ester bonds yields one fatty-acid salt.
Molar mass of $C_{17} H_{35} COONa$ . This formula just groups the tail ( $C_{17} H_{35}$ ) with the salt head ( $COONa$ ). Add up all the atoms in one place — $C$ : $17 + 1 = 18$ , $H$ : $35$ , $O$ : $2$ , $Na$ : $1$ — so it is the same molecule written $C_{18} H_{35} O_{2} Na$ . Mass $= 18 (12) + 35 (1) + 2 (16) + 23 = 216 + 35 + 32 + 23 = 306 g/mol$ . Why this step? We need grams; grams = moles × molar mass, and both formulas give the same 306.
Mass of soap. $1.5 mol \times 306 g/mol = 459 g$ .

Verify: Sanity-check the stoichiometry that produced 459 g: $0.5 mol$ fat needs $0.5 \times 3 = 1.5 mol$ NaOH ( $1.5 \times 40 = 60 g$ ) and yields $1.5 mol$ soap and $0.5 mol$ glycerol. Mass in $= 445 + 60 = 505 g$ ; mass out $= (1.5 \times 306) + (0.5 \times 92) = 459 + 46 = 505 g$ . Mass balances exactly, confirming the answer 459 g. ✓

Cell D — Degenerate input: a monoglyceride (only 1 ester)

Worked example Example 4 (Cell D)

A monoglyceride = glycerol carrying just one fatty acid (the other two –OH are free). Saponify one mole of it. How many moles of soap and how many free –OH remain on the released glycerol?

Forecast: how does the "3" from Example 1 shrink?

Count esters present. Only 1 ester bond (2 of glycerol's 3 –OH never reacted). Why this step? You can only break bonds that exist — this is the degenerate/edge case where the usual "3" collapses.
Saponify. 1 ester + 1 NaOH → 1 mol soap + glycerol. Why this step? Same rule as before, just with 1 instead of 3.
Free –OH on glycerol product. Glycerol always regenerates with all 3 –OH free (2 were already free, the 3rd is freed by hydrolysis).

Verify: For a diglyceride you'd get 2 soap; triglyceride 3. The soap count = ester count, and the released glycerol is always the same molecule with 3 –OH. Pattern holds across mono/di/tri. ✓

Cell E — Base swap: NaOH vs KOH

Worked example Example 5 (Cell E)

The same fat is saponified once with NaOH and once with KOH. (a) Which mass of base is needed per mole of ester? (b) Which gives "hard" bar soap vs "soft" liquid soap?

Forecast: which metal makes the harder soap?

Molar masses of the bases. $NaOH = 23 + 16 + 1 = 40 g/mol$ ; $KOH = 39 + 16 + 1 = 56 g/mol$ . Why this step? Same stoichiometry (1 base per ester), so the only difference is the metal mass.
(a) Per ester: 40 g NaOH or 56 g KOH. KOH is heavier per mole.
(b) Sodium salts pack tightly into a rigid lattice → hard bar soap. Potassium ions are larger, pack loosely → soft/liquid soap. Why this step? This is the Soaps and Detergents structure→property switch: cation size sets crystallinity.

Verify: $56/40 = 1.4$ — KOH always weighs 40 % more per mole than NaOH. And the "K = soft, Na = hard" rule matches real potash (liquid) vs caustic-soda (bar) soaps. ✓

Cell F — Acid hydrolysis vs saponification (product switch)

Worked example Example 6 (Cell F)

The same triglyceride is hydrolysed two ways: (i) with dilute $H^{+}$ + water, (ii) with NaOH. Write the fatty-acid product of each and say which reaction runs to completion. (Recall $R$ = the long hydrocarbon tail.)

Forecast: free acid or a salt — for each medium?

Acid path (i). Water splits the ester; the fatty acid leaves as the free acid $R-COOH$ . Reaction is an equilibrium — reversible. Why this step? The acid product can re-esterify with glycerol, so it never fully finishes.
Base path (ii). NaOH splits the ester and neutralises the acid → sodium salt $R-COO^{-} Na^{+}$ (soap). Why this step? The salt cannot re-esterify → the equilibrium is trapped forward → goes to completion.
The switch. Medium sign: acid → free $R-COOH$ (reversible); base → $R-COO^{-} Na^{+}$ (complete).

Verify: This is exactly the common-mistake fix from the parent note: soap requires base because only the salt "locks in" the products. Same bond broken, different product because of the medium. ✓

Cell G — Amphipathic geometry (word problem, with figure)

Worked example Example 7 (Cell G)

You drop phospholipids into a beaker of pure water. Which structure forms, and which way do the phosphate heads and hydrocarbon tails point? Then: you dissolve a drop of oil first — now what forms around the oil?

Forecast: bilayer? sphere? Sketch it in your head before looking.

Identify the two personalities. Phosphate head = hydrophilic (loves water); two fatty tails = hydrophobic (hate water). This dual nature is amphipathic. Why this step? The geometry is decided entirely by which end can touch water.
In pure water → bilayer. Tails hide inward facing each other; heads face water on both faces (left figure). This sheet is the cell membrane. Why this step? A sheet lets every tail escape water while every head stays wet — the stable compromise.
With an oil drop → micelle. Tails point inward into the oil, heads point outward into water, wrapping the grease in a ball (right figure). Why this step? Now the tails have somewhere non-water to hide (the oil), so a single-layer sphere suffices — this is exactly how soap lifts grease.

Verify: In both cases the rule "tails avoid water, heads meet water" is obeyed — bilayer when there's only water, micelle when there's a greasy core. Same physics as saponified soap cleaning. ✓

Cell H — Limiting case: fully unsaturated / metabolic fate

Worked example Example 8 (Cell H)

Push unsaturation to the limit: compare a highly unsaturated oil ( $18 : 3$ ) with a saturated fat ( $18 : 0$ ) for (a) physical state and (b) how much energy per gram on oxidation. Which is the denser fuel?

Forecast: does adding double bonds change the energy stored?

State at the limit. Many kinks ( $d = 3$ ) → cannot pack at all → liquid at low temperature (from Example 2's trend extended). Why this step? Confirms the melting-point switch behaves smoothly as $d$ grows — no surprise scenario left.
Energy density. Both are ~18-carbon chains rich in C–H bonds. Full oxidation via beta-oxidation gives ≈ 9 kcal/g — far above carbohydrate's ≈ 4 kcal/g. Why this step? Energy comes from reduced C–H bonds; a double bond removes 2 H, so the unsaturated chain stores slightly less energy per gram, but both crush carbohydrate.
Denser fuel. The saturated $18 : 0$ (more C–H bonds, no C=C) stores marginally more energy per gram than $18 : 3$ .

Verify: $9/4 = 2.25$ — fat carries just over twice the energy per gram of carbohydrate, and unsaturation only nudges this slightly down. Limiting behaviour (max kinks) gives max fluidity, minimal packing — consistent with everything above. ✓

Cell I — Exam twist: iodine number

Worked example Example 9 (Cell I)

Iodine number = grams of $I_{2}$ that add across the $C = C$ double bonds of 100 g of fat (exactly one $I_{2}$ molecule adds per double bond). Oil P has iodine number 0; oil Q, made of pure oleic acid ( $18 : 1$ , $M = 282 g/mol$ ), has some larger value. (a) Which oil is saturated? (b) Compute the iodine number of oil Q, and (c) explain how this single number measures $d$ . Use $M_{I_{2}} = 254 g/mol$ .

Forecast: does a bigger iodine number mean more or fewer double bonds?

What $I_{2}$ reacts with. Iodine adds only across $C = C$ double bonds — one $I_{2}$ molecule per double bond — and does not touch the rest of the chain. Why this step? This turns an invisible count of double bonds into a weighable mass, giving us a chemical "ruler" for $d$ .
(a) Interpret P. Iodine number 0 → absorbs no iodine → no double bonds → $d = 0$ → saturated (a straight, solid fat). Why this step? Zero uptake is only possible if there is nothing for iodine to add to — the degenerate lower end of the scale.
(b) Moles of double bonds in 100 g of oil Q. Oleic acid is $18 : 1$ , so each molecule has exactly 1 double bond. Moles of oil $= \frac{100 g}{282 g/mol} = 0.355 mol$ , and since $d = 1$ that is also $0.355 mol$ of double bonds. Why this step? Iodine uptake counts double bonds, so we must first count moles of double bonds, not moles of molecules in general.
(b, cont.) Mass of $I_{2}$ absorbed = iodine number. $0.355 mol \times 254 g/mol = 90 g$ of $I_{2}$ per 100 g of oil → iodine number ≈ 90. Why this step? By definition the iodine number is the grams of $I_{2}$ absorbed by 100 g of fat, so this mass is the answer directly.
(c) How the number measures $d$ . For a fixed chain length, more double bonds → more $I_{2}$ added → higher iodine number. So iodine number rises with $d$ : 0 for saturated P, ≈90 for mono-unsaturated Q, and even higher for poly-unsaturated oils. It ranks fats by unsaturation with a simple weighing — no spectroscopy needed. Why this step? This is the exam payoff: one measurable mass replaces a hidden structural count.

Verify: $100/282 = 0.3546 mol$ double bonds; $0.3546 \times 254 = 90.1 g$ of $I_{2}$ → iodine number ≈ 90 (nonzero, large), while saturated P sits at 0. This confirms "more double bonds → higher iodine number." ✓

Recall Self-test: name the cell

Soap mass from a given fat mass — which cell? ::: Cell C (molar mass / yield) Why base makes soap go to completion but acid does not? ::: Cell F (product switch: salt locks the equilibrium) What forms when phospholipids meet an oil drop? ::: Cell G — a micelle (tails into oil, heads into water) A monoglyceride uses how many NaOH to saponify? ::: Cell D — just 1 (only 1 ester bond) Iodine number 0 means the fat is what? ::: Cell I — fully saturated ( $d = 0$ )